# Why is "time orthogonal to space" in inertial reference frames?

• B
Here is an exercise that may help to clarify the above statements of mine.

Consider Rindler coordinates . This is a different coordinate chart on Minkowski spacetime, in which the metric of Minkowski spacetime has a different representation that is not $\text{diag}(-1,1,1,1)$.

However, this coordinate chart is non-inertial. Try showing this by showing that this chart does not meet either of the two requirements for an IRF: free particles do not have constant coordinate velocity, and light does not always have speed $c$.

 https://en.wikipedia.org/wiki/Rindler_coordinates
One final thing: what mathematical background would one need to prove the result we want to (regarding the orthogonality requirement of an IRF)? Would linear algebra (up to basic knowledge of covectors / vectors and transformation properties of tensors) suffice? Or would it require more advanced stuff like differential geometry and require me to know fancy concepts like Ricci curvature, etc.? I'll accordingly brush up on the needed mathematical prerequisites.

PeroK
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@Shirish I won't interrupt your thread any more. I was hoping my questions would help you, which I why I asked them. I would have hoped you could learn SR without getting into these deep conceptual waters. From a pedagogical point of view I think this treatment is asking a lot. Rindler coordinates, Riemann tensor, coordinate-free descriptions of spacetime. These are things you need to wrestle with eventually. But, up front I think it's a lot to ask of the student.

• Shirish
robphy
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Without a metric you cannot define timelike or spacelike or null at all for vectors. See below.
Yes, again I'm a little sloppy.

I was going to write a more complete reply...
but saw that it takes too long to nail down the ideas, and will go off-topic for this thread.

I'll stand by my intuitive characterization
that
"spacelike" only has meaning after a metric is introduced,
in the sense that
given an observer's worldline on a position-vs-time graph,
the direction that is "purely spatial" to it
is determined by a metric [say, determined by experiment]
as the tangent line to a conic that describes
the set of point-events equidistant from the origin.

Then
the set of proper metric-preserving transformations will
lead to characterizing a subset of vectors to be "spacelike" from such purely-spatial vectors.

For more details of this viewpoint (still under development),

@Shirish I won't interrupt your thread any more. I was hoping my questions would help you, which I why I asked them. I would have hoped you could learn SR without getting into these deep conceptual waters. From a pedagogical point of view I think this treatment is asking a lot. Rindler coordinates, Riemann tensor, coordiante-free descriptions of spacetime. These are things you need to wrestle with eventually. But, up front I think it's a lot to ask of the student.
Oh no you weren't interrupting, no worries It's a chance for all of us (especially me haha) to clarify concepts. I want to go into deep conceptual waters sooner or later for sure. I might be getting ahead of myself at times and that might slow me down, but I guess that's a downside of self-teaching.

• PeroK
PeterDonis
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what mathematical background would one need to prove the result we want to (regarding the orthogonality requirement of an IRF)?
Ordinary algebra.

Work in 1+1 spacetime for simplicity, as you suggested.

Since we already have one coordinate chart that satisfies the requirements, we can re-formulate our question as follows: what transformations could we make from this chart to some other chart, that would preserve the two properties required for an IRF (free particle worldlines are straight lines, and light always has speed $c$)? What constraints can we show must hold for such transformations?

If you find that the required constraints narrow down all of the possible transformations to just the Lorentz transformations in the standard form for the Minkowski chart, you're done: you've proven that only one particular kind of coordinate chart, the standard $(t, x)$ Minkowski chart in which the metric is $\text{diag}(-1, 1)$, will work, since the standard Lorentz transformations starting from such a chart can only take you to the same kind of chart.

The constraints that can be made on the transformations from the two IRF requirements can be expressed using just ordinary algebra.

• Shirish
PeterDonis
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I'll stand by my intuitive characterization
that
"spacelike" only has meaning after a metric is introduced,
This was not in dispute. I agree with it.

given an observer's worldline on a position-vs-time graph,
the direction that is "purely spatial" to it
is determined by a metric [say, determined by experiment]
as the tangent line to a conic that describes
the set of point-events equidistant from the origin.

Then
the set of proper metric-preserving transformations will
lead to characterizing a subset of vectors to be "spacelike" from such purely-spatial vectors.
This, however, appears to be based on the reference you linked to, which is, as you note, still under development, and is off topic for this thread. (The reference in general is beyond a "B" level discussion anyway; it's at least "I" and possibly "A".)

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Since we already have one coordinate chart that satisfies the requirements, we can re-formulate our question as follows: what transformations could we make from this chart to some other chart, that would preserve the two properties required for an IRF (free particle worldlines are straight lines, and light always has speed $c$)? What constraints can we show must hold for such transformations?

If you find that the required constraints narrow down all of the possible transformations to just the Lorentz transformations in the standard form for the Minkowski chart, you're done
So this is something I managed to do assuming that the transformation from chart $(t,x)$ to $(t',x')$ is linear, constancy of light speed and isotropy/homogeneity of space. We know that in chart $(t,x)$, by construction, the time and space basis vectors are orthogonal (equipped with $\text{diag}(-1,1,1,1)$ metric). Again because of the way we constructed it, free particles have straight worldlines and light has 45 degree worldlines.

(Gap in proof: To prove that the chart $(t',x')$ also has the $\text{diag}(-1,1,1,1)$ metric). Once we know that even the new chart $(t',x')$ has $\text{diag}(-1,1,1,1)$ metric (say $M$), then we know that under a linear transformation, $v_1^TMv_2$ is invariant. Also, the timelike basis vector in the old chart will be mapped to a timelike basis vector in the new one, and similarly for spacelike basis vector (again assuming $\text{diag}(-1,1,1,1)$ metric for the new chart). So
$$0 = \mathbf{e}_t^T M \mathbf{e}_x = \mathbf{e}_t^T L^T M L \mathbf{e}_x = \mathbf{e'}_t^T M \mathbf{e'}_x$$
since $L^TML=M$, which proves orthogonality of timelike and spacelike basis vectors in the new chart as well.

So I guess this shows that the only transformation mapping IRFs to IRFs, i.e. the Lorentz transformation, preserves the orthogonality between timelike and spacelike vectors. The only gap is to show that the chart $(t',x')$ also has the $\text{diag}(-1,1,1,1)$ metric.

PeterDonis
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assuming that the transformation from chart $(t,x)$ to $(t',x')$ is linear, constancy of light speed and isotropy/homogeneity of space
The first (linear) shouldn't be an assumption; you should be able to prove that it's required based on the requirements of an IRF.

The second is fine since it's a requirement of an IRF.

The third shouldn't be an assumption either; you should also be able to prove that it's required based on the requirements of an IRF.

Without proofs of the first and third assumptions based on the requirements of an IRF, what you are doing is not proving anything.

PeterDonis
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the timelike basis vector in the old chart will be mapped to a timelike basis vector in the new one
No, it won't. This is certainly not true for a standard Lorentz transformation from one standard Minkowski chart to another. (If you had left out the word "basis" the second time you used it, your statement would be correct; but it's false with the second "basis" included.)

similarly for spacelike basis vector
Same remark as above: basis vectors in the old chart will not be basis vectors in the new chart. Their signature (timelike or spacelike) and norm will be preserved, but not the fact of them being basis vectors.

The first (linear) shouldn't be an assumption; you should be able to prove that it's required based on the requirements of an IRF.

The second is fine since it's a requirement of an IRF.

The third shouldn't be an assumption either; you should also be able to prove that it's required based on the requirements of an IRF.

Without proofs of the first and third assumptions based on the requirements of an IRF, what you are doing is not proving anything.
Linearity of the transformation follows from the fact that in IRFs free particles have constant speed. The chart $(t,x)$ is rectilinear, so the free particle worldline is straight in it. Assuming that the new chart $(t',x')$ representing another IRF is also rectilinear, that particle's worldline will still be straight due to constant speed. Map from straight to straight worldlines implies linearity. (I'm not sure what we can do if we don't assume that the chart $(t',x')$ is rectilinear.

Third point: There's no preferred location in space nor a preferred direction, no matter where we are or which direction we're facing, we shouldn't be able to distinguish between IRFs, since all IRFs are assumed to be physically equivalent via the principle of relativity. So hopefully this is an acceptable physical justification of the homogeneity and isotropy assumptions.

(If you had left out the word "basis" the second time you used it, your statement would be correct; but it's false with the second "basis" included.)

Same remark as above: basis vectors in the old chart will not be basis vectors in the new chart. Their signature (timelike or spacelike) and norm will be preserved, but not the fact of them being basis vectors.
Yes, my mistake. I should've said that spacelike (timelike) basis vectors are mapped to spacelike (timelike) basis vectors in the new chart. But this puts another hole in my argument - now I can only say that the images of the old chart's basis vectors are orthogonal. This doesn't enable me to say anything about the new chart's basis vectors - whether both have the same or different signatures, or the inner product of those. pervect
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Experimentally, time being orthogonal to space is the same as saying that we use Einstein clock synchronization convention, as a way to get "fair" or "isotropic" clock synchronization.

When we make time orthogonal to space, physics work the way we usually expect. Light moves the same in both directions, and so do electron beams with the same energy. It doesn't matter to the physics which way the light or the electron beam is going, it's speed is the same.

If we don't make time orthogonal to space, both the speed of light and the speed of electron beams of a specified energy depends on the direction. This is most obvious for high energy electron beams, as they approach the speed of light as one increases their energy per electron. Thus if the behavior of light is not isotropic, neither is the behavior of other things. See for instance and/or the peer reviewed paper associated with the video. The general topic is known as "isotropy". Sometimes people appear to think isotropy applies only to light, but it applies to other physical phenomenon as well.

If the idea that the speed of identical objects (identical because they have a known energy, or momentum) moving in different directitons is uncomfortable, then one should stick with isotropic coordinates and/or inertial frames, which implies that time must be orthogonal to space.

PeterDonis
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Linearity of the transformation follows from the fact that in IRFs free particles have constant speed.
Yes.

Assuming that the new chart $(t',x')$ representing another IRF is also rectilinear
I don't think this assumption for the new chart is actually required. If you look at coordinate speeds in non-rectilinear charts, you will see that they can't possibly be constant in all directions. For example, consider $d\theta / dt$ for a free particle in spherical coordinates (and ignore the edge case where the particle happens to be moving along an axis that makes $\theta$ constant). So constant coordinate speed in all directions already implies a rectilinear chart.

hopefully this is an acceptable physical justification of the homogeneity and isotropy assumptions.
It's a good start, but you spoke in general terms. There is a specific requirement for an IRF that requires isotropy.

I should've said that spacelike (timelike) basis vectors are mapped to spacelike (timelike) basis vectors in the new chart.
Meaning, mapped by the coordinate transformation? Yes, you can think of the transformation that way, but you can also think of it as simply changing the components of vectors, but not the vectors themselves. In other words, the old basis vectors will have more than one nonzero component in the new chart, and the new basis vectors will have more than one nonzero component in the old chart. The question then is how the coordinate transformation can change those components while still preserving the required properties for an IRF.

now I can only say that the images of the old chart's basis vectors are orthogonal.
Orthogonality is a coordinate-independent property, so a pair of vectors that are orthogonal in one chart will be orthogonal in any chart.

This doesn't enable me to say anything about the new chart's basis vectors - whether both have the same or different signatures, or the inner product of those.
You can't assume the new basis vectors are orthogonal, or even that one is timelike and three are spacelike, that's true. You have to prove that, if it is provable.

You do have information about the new basis vectors, though: you know that free particles still have to have straight worldlines, and that light still has to have speed $c$. And you know that the coordinate transformation has to be linear. So you do have information to work from.

• Shirish
...you know that free particles still have to have straight worldlines, and that light still has to have speed $c$. And you know that the coordinate transformation has to be linear. So you do have information to work from.
This info is used to derive the Lorentz transformation - in the $(t,x)\to(t',x')$ case, it's $\mathcal{L}=\begin{bmatrix}\gamma & \beta\gamma\\\beta\gamma & \gamma\end{bmatrix}$ (sign of off-diagonal terms depends on relative velocity so let's consider them positive in this case).

Orthogonality is a coordinate-independent property, so a pair of vectors that are orthogonal in one chart will be orthogonal in any chart.
Now this is important info and something I'm embarrassed to admit that I didn't know. I thought if orthogonality is coordinate-independent as you said, then maybe even the inner product is (googling confirmed this). So if we have a couple of vectors $\mathbf{a},\mathbf{b}$ with representations $[ a ]_{\mathcal{A}}, [ b ]_{\mathcal{A}}$ in the initial coordinate chart $\mathcal{A}$ (and since we know that $\eta=\text{diag}(-1,1)$ is the metric representation in $\mathcal{A}$ by construction), their inner product is $[ a ]^T_{\mathcal{A}}\eta[ b ]_{\mathcal{A}}$.

Let $\eta'$ be the metric representation in the new coordinate chart $\mathcal{B}$ (the Minkowski metric doesn't change - its representation changes - something I realized after reading your post and googling related stuff). The representation of $\mathbf{a}$ in $\mathcal{B}$ will be $[ a ]_{\mathcal{B}}=\mathcal{L}[ a ]_{\mathcal{A}}$ (similarly for $\mathbf{b}$). Then
$$[ a ]^T_{\mathcal{B}}\eta'[ b ]_{\mathcal{B}}=[ a ]^T_{\mathcal{A}}\eta[ b ]_{\mathcal{A}}=[ a ]^T_{\mathcal{A}}\mathcal{L}^T(\mathcal{L}^T)^{-1}\eta\mathcal{L}^{-1}\mathcal{L}[ b ]_{\mathcal{A}} =[ a ]^T_{\mathcal{B}}(\mathcal{L}^{-1})^T\eta\mathcal{L}^{-1}[ b ]_{\mathcal{B}} \\\implies \eta'=(\mathcal{L}^{-1})^T\eta\mathcal{L}^{-1}=\eta$$
So this was the missing link. And now since the metric representation is the same even in the new coordinate chart, and since the representation of the new basis vectors in the new chart $\mathcal{B}$ is $[1,0]$ and $[0,1]$, their inner product is zero.

It's a good start, but you spoke in general terms. There is a specific requirement for an IRF that requires isotropy.
I confess I don't know about that requirement. As far as I've read, isotropy is justified loosely by the fact that the orientation of the relative velocity vector between IRFs can be arbitrary. If there's an even stronger/more specific justification, I'm lost.

PeroK
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Orthogonality is a coordinate-independent property, so a pair of vectors that are orthogonal in one chart will be orthogonal in any chart.
Although that itself does need to be proved at some stage.

• Shirish
PeterDonis
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This info is used to derive the Lorentz transformation
How?

I thought if orthogonality is coordinate-independent as you said, then maybe even the inner product is
It is.

the metric representation is the same even in the new coordinate chart
It is if you use the Lorentz transformation, yes. But you still have to show how you derive the Lorentz transformation given the info you quoted at the top of your post (see my first question above).

I confess I don't know about that requirement.
Sure you do; we've mentioned it repeatedly in this discussion. It's even mentioned in the first quote from my previous post that you give in this post: look at the clause after the comma in the first sentence of that quote.

PeterDonis
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that itself does need to be proved at some stage
It's obvious from the definition of orthogonality and the fact that inner products are coordinate-independent. And before you say that inner products being coordinate-independent has to be proved, that's part of the definition of a metric.

• Shirish
But you still have to show how you derive the Lorentz transformation given the info you quoted at the top of your post (see my first question above).
One thing - assuming that the transformation is Lorentz, does the argument in my post (which shows $\eta=\eta'$ and hence orthogonality of basis in the new chart) seem okay to you?

As for Lorentz, there's a related thread here that I'd created previously in which I showed steps for showing that the transformation is Lorentz: https://www.physicsforums.com/threa...locks-are-synchronized-at-each-origin.989150/

The proof in the top post in that thread is incomplete, but folks at PF helped me complete it using homogeneity and isotropy assumptions (homogeneity to show that without loss of generality, we can assume a linear transformation instead of affine, and isotropy to show that $\gamma(v)$ is an even function of $v$).

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PeterDonis
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assuming that the transformation is Lorentz, does the argument in my post (which shows $\eta=\eta'$ and hence orthogonality of basis in the new chart) seem okay to you?
If the transformation is Lorentz, then yes, you're just expressing the well-known fact that the Lorentz transformation preserves the Minkowski form $\eta$ of the metric.

there's a related thread here that I'd created previously in which I showed steps for showing that the transformation is Lorentz
Ah, ok. Yes, if you use the derivation given in that thread, you're basically showing that the transformation has to be Lorentz in order to satisfy the speed of light being $c$ in the new coordinates, which is one of the requirements for an IRF.

If the transformation is Lorentz, then yes, you're just expressing the well-known fact that the Lorentz transformation preserves the Minkowski form $\eta$ of the metric.

Ah, ok. Yes, if you use the derivation given in that thread, you're basically showing that the transformation has to be Lorentz in order to satisfy the speed of light being $c$ in the new coordinates, which is one of the requirements for an IRF.
Yay! So finally my mind is at peace. Now I can happily continue with the rest of the studies (until next time a doubt derails me again )

Thank you so much

• berkeman
PeterDonis
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Thank you so much
You're welcome!